So, the speed of gravity is the same as the speed of light, yes? So that mean, supposedly in the solar system, I suddenly put a reasonably sized black hole in the center of the sun, pluto would be affected by the force of gravity about 8 hours later right?

More like 4-7 hours, but yes. (And this is assuming you "put a black hole in the sun" by adding a large dense mass to it, not just by compressing the sun's existing mass, since the latter would have almost no effect at all at that distance).

That's ripples, not evaporation of mass (don't confuse things like this with Hawking Radiation). A finger (or two) whisked around in water, rather than drawing up (or dumping further in) water or (more accurately) the displacing less or more water with the hands.

Anyway, a solar-sized black-hole will feel the same as a solar-sized sun (gravitationally) out there. The process of change might ripple space-time (depending on what hapoens to make the swap or transition), but the bits of the black-hole's gravity gradient that 'feel' different to the Sun's gradient are already where you wouldn't normally want to be with the Sun there...

wikipedia wrote:The energy released by the binary as it spiralled together and merged was immense, with the energy of 3.0+0.5−0.5 c2 solar masses (5.3+0.9−0.8×1047 joules or 5300+900−800 foes) in total radiated as gravitational waves

wikipedia wrote:The energy released by the binary as it spiralled together and merged was immense, with the energy of 3.0+0.5−0.5 c2 solar masses (5.3+0.9−0.8×1047 joules or 5300+900−800 foes) in total radiated as gravitational waves

See what you mean, true, true. Even though the energy radiated still has mass, it is being spread out very rapidly, essentially "vanishing" by some number of definitions.

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Its worth noting though that the speed of light and gravity is no faster or slower than causality, there is no "waiting for gravity" because you cannot ever know of an event in advance of the lightspeed event-front. Gravitational effects will always be noticed at the exact same time as the event that caused them. Calculating different "times" for events seperated by astronomical distances has far less meaning than doing something similar in a non-relativistic, macro-scale context.

andykhang wrote:Oh thanks. So when calculating the effect of gravity at astronomical range, you must also take this in consideration right?

Yes. Say for example you had some very large mass somewhere a long way away, like a black hole, and something somehow made it suddenly move quickly in your direction, and then back again. As it moved your direction, its gravitational effects on you would increase (by a very tiny, hard-to-detect amount), and as it moved away they would go back as before, and if you had sensitive enough equipment you could see something moving (or distorting along with the changing spacetime, at least) in your lab in response to that distant movement of large masses. But not until enough time had passed for the change in gravity to reach you, which would be the same amount of time that light from the event would take to reach you.

If you had distant large masses moving back and forth rapidly, like say, two black holes or neutron stars orbiting each other extremely rapidly as they merged, then your super-sensitive lab equipment could measure the "waves" (if you will) of changing gravity that were radiated by that event, and those "gravity waves" would travel at the speed of light.

Tangential, but to do with gravitational waves: what sort of masses would need to collide to produce waves that would be biologically relevant (nanometer scale, say) from one AU out and/or the distance to proxima centauri?

That said, the strain due to gravitational radiation falls off as r-2, right? So if the black hole merger 400 Mpc away produced an observed strain on Earth of 2.5 x 10-22, then a merger of similar size at 1 AU should produce a strain of 1.6 x 106 if my calculations are correct . . . which doesn't seem plausible. How am I actually supposed to calculate this?

EDIT: The strain must fall off as r-1, which explains how it is possible to detect such distant events. That would correspond to a strain at 1 AU of 2.1 x 10-8, or about 37 nm of height change for a 1.8 m human, which is not only survivable but completely negligible. However, it matches the "nanometer-size" limit you wanted to use, so by that metric, two black holes of about 30 and 35 solar masses colliding at a distance of 1 AU should do the trick.

At Proxima Centauri, a distance of 1.3 pc = 2.7 * 105 AU, we would need a source 2.7 * 105 times as powerful to get an equivalent strain on Earth.

The 'waves' are from the gravity one feels, outwith the event horizon. They are waves because the mass (the two masses) is(/are) moving through space to oscillate the effect of the combined gravitational gradient for the observer.

And you might say that gravity itself must 'escape' the event horizon, to get a sense of the mass and gravity of which the black hole is formed, though that might only be the equivalent of shell-theorem. All gravity might as well be at the event horizon. And it is that shell (or combination of shells, where multiple objects interact, then merge) that is the edge of the "finger in the pond" beng whisked around, the core singularity doesn't touch observable space like the phalangeal bones of the hands don't ever (one hopes!) directly touch the water.

If (big if!) some similar form of ripple-worthy gravitational disturbance existed betwixt EH and Singularity, would it actually matter (NPI) to anything outside the horizon, as you'd only 'feel' the sum mass (and charge, and angular momentum?), not any of the details. It would violate the concept of not being able to 'see' within the forbidden zone. (Or, if not so no-hair, it might be the answer to mapping its depths, like seismic waves help map our Earth's interior. But I'm putting that in my "I doubt it" bucket.)

Soupspoon wrote:The 'waves' are from the gravity one feels, outwith the event horizon. They are waves because the mass (the two masses) is(/are) moving through space to oscillate the effect of the combined gravitational gradient for the observer.

And you might say that gravity itself must 'escape' the event horizon, to get a sense of the mass and gravity of which the black hole is formed, though that might only be the equivalent of shell-theorem. All gravity might as well be at the event horizon. And it is that shell (or combination of shells, where multiple objects interact, then merge) that is the edge of the "finger in the pond" beng whisked around, the core singularity doesn't touch observable space like the phalangeal bones of the hands don't ever (one hopes!) directly touch the water.

If (big if!) some similar form of ripple-worthy gravitational disturbance existed betwixt EH and Singularity, would it actually matter (NPI) to anything outside the horizon, as you'd only 'feel' the sum mass (and charge, and angular momentum?), not any of the details. It would violate the concept of not being able to 'see' within the forbidden zone. (Or, if not so no-hair, it might be the answer to mapping its depths, like seismic waves help map our Earth's interior. But I'm putting that in my "I doubt it" bucket.)

I was of the impression that gravity waves were somewhat more involved than merely the change in gravity as a large mass oscillates further and closer, that it takes an accelerating mass, not just a moving one, to produce them?

Merely moving, presumably past an observer/vice-versa, shouldn't do the water thing of creating a wake of waves, because of relativity. (If you have a bow-wave, you know you're the one moving, w.r.t. the medium!)

So take the water-whisking with a pinch of sea-salt, if you want. It's just an easier initial analogy than the marginally better infinite frictionless rubber sheet.

Well two neutron stars orbiting each other have no means of slowing down and spiralling-in, other than they lose energy via gravity wave radiation, so do then ALL rotating masses lose energy in this way (just a vanishingly small amount for "normal" sized masses/velocities) ?

I think the confusion lies with things that are accelerating being capable of (and were) described as 'merely' moving, even though not everything that is moving (and nothing with constant velocity) is accelerating.

(Again with relativity, if the subjective observer sees it moving straight around curved spacetime. Or not moving, just freefalling, as the rest of the universe wheels around you.)

Wikipedia's article includes an image of a circle of points being affected by regular gravitational waves.

My understanding is that if gravitational strain is static, it's like one frame of the animation, so particles are just in a different position but they don't move. If it's changing linearly, then the particles just keep moving with an apparent initial velocity. If the second derivative is nonzero, though, then the particles move in a nonlinear way.

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Well two neutron stars orbiting each other have no means of slowing down and spiralling-in, other than they lose energy via gravity wave radiation, so do then ALL rotating masses lose energy in this way (just a vanishingly small amount for "normal" sized masses/velocities) ?

A single axially symmetric rotating body doesn't radiate gravitational waves, but a pair of bodies in orbit does.

This pdf has a lot of stuff that's way beyond me, but on page 58 (of the file, p. 1238 of the book it's from) it gives an equation for how the period changes for a pair of equal masses in circular orbit: -3.4e-12 M^(5/3) P^(-5/3)where M is in solar masses and P is the orbital period in hours. (The resulting number is dP/dt, so dimensionless, as in "seconds of period change per second")

A pair of solar masses in a 1000-hour orbit (chosen to make the 5/3 power simple) would spiral in at a rate of 3.4e-17, or 3.4e-14 hours per orbit, or about 1 nanosecond per year. So for normal binaries losing a second per (several) billion years or so is going to be dwarfed by just about every single other thing that might happen over the course of a star's life.

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p1t1o wrote:The gravity from the pair that I experience is equivalent to 2 solar masses sitting at the centre-of-mass of the binary pair, correct?

No.

That's a good approximation if you're much farther out than the distance between the stars, but it isn't exact because the closer star has more of an effect including when it's off-center.

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